Question

Prove that the points (–7, –3), (5, 10), (15, 8) and (3, –5) taken in order are the corners of a parallelogram.

Answer 1

proved.
just wanted to help.
reply if it is correct

Answer 2

Answer:

Step-by-step explanation:

From the figure, AB=$\sqrt{(5+7)^{2}+(10+3)^{2}}$

=$\sqrt{144+169}$

AB=$\sqrt{313}$

BC=$\sqrt{(15-5)^{2}+(8-10)^{2}}$

=$\sqrt{100+4}$

BC=$\sqrt{104}$

CD=$\sqrt{(15-3)^{2}+(8+5)^{2}}$

=$\sqrt{144+169}$

CD=$\sqrt{313}$

DA=$\sqrt{(3+7)^{2}+(-5+3)^{2}}$

=$\sqrt{100+4}$

DA=$\sqrt{104}$

Now, AB=CD=$\sqrt{313}$ and BC=DA=$\sqrt{104}$, therefore opposite sides are equal. Hence, This is a parallelogram.